题目：给定一个头结点，倒叙输出一个链表

解法1：先将链表反转，在遍历输出

解法2：不修改链表自身的结构，动态申请一段空间，申请一个指针数组，数组内存放的指针指向链表的每个值，再遍历数组输出：

void PrintListBack(ListNode* head){	int n = GetLength(head);	ListNode** p = new ListNode*[n+1];	p[n] = NULL;	int i = 0;	ListNode* pNode = head;	while (i < n)	{		p[i] = pNode;		pNode = pNode->m_pNext;		++i;		}	for (i = n-1; i>= 0; i--)	{		cout<<p[i]->m_nValue<<" ";	}}

解法3：利用栈，后进先出；

void PrintListBack2(ListNode* head){	stack<ListNode> listStack;	ListNode* pNode = head;	while (pNode!=NULL)	{		listStack.push(pNode->m_nValue);		pNode = pNode->m_pNext;	}	int i = listStack.size();	for (;i>0;i--)	{		ListNode p = listStack.top();		cout<<p.m_nValue<<endl;		listStack.pop();	}}void PrintListBack3(ListNode* head){	stack<ListNode*> listStack;	ListNode* pNode = head;	while (pNode!=NULL)	{		listStack.push(pNode);		pNode = pNode->m_pNext;	}	int i = listStack.size();	for (;i>0;i--)	{		ListNode* p = listStack.top();		cout<<p->m_nValue<<endl;		listStack.pop();	}}

解法4：利用递归

void PrintListBack4(ListNode* head){	if (head != NULL)	{		if (head->m_pNext != NULL)		{			PrintListBack4(head->m_pNext);		}		cout<<head->m_nValue<<endl;	}}

完整代码：

// PrintListBack.cpp : 定义控制台应用程序的入口点。//#include "stdafx.h"#include <iostream>#include <stack>using namespace std;struct ListNode {	int        m_nValue;	ListNode*  m_pNext;	ListNode(int a):m_nValue(a),m_pNext(NULL)	{	}};int GetLength(ListNode* head){	ListNode* pNode = head;	int nCount = 0;	while (pNode!= NULL)	{		nCount++;		pNode = pNode->m_pNext;	}	return nCount;}void PrintListBack(ListNode* head){	int n = GetLength(head);	ListNode** p = new ListNode*[n+1];	p[n] = NULL;	int i = 0;	ListNode* pNode = head;	while (i < n)	{		p[i] = pNode;		pNode = pNode->m_pNext;		++i;		}	for (i = n-1; i>= 0; i--)	{		cout<<p[i]->m_nValue<<" ";	}}void PrintListBack2(ListNode* head){	stack<ListNode> listStack;	ListNode* pNode = head;	while (pNode!=NULL)	{		listStack.push(pNode->m_nValue);		pNode = pNode->m_pNext;	}	int i = listStack.size();	for (;i>0;i--)	{		ListNode p = listStack.top();		cout<<p.m_nValue<<endl;		listStack.pop();	}}void PrintListBack3(ListNode* head){	stack<ListNode*> listStack;	ListNode* pNode = head;	while (pNode!=NULL)	{		listStack.push(pNode);		pNode = pNode->m_pNext;	}	int i = listStack.size();	for (;i>0;i--)	{		ListNode* p = listStack.top();		cout<<p->m_nValue<<endl;		listStack.pop();	}}void PrintListBack4(ListNode* head){	if (head != NULL)	{		if (head->m_pNext != NULL)		{			PrintListBack4(head->m_pNext);		}		cout<<head->m_nValue<<endl;	}}int _tmain(int argc, _TCHAR* argv[]){	ListNode*  head  = new ListNode(0);	ListNode*  node1 = new ListNode(1);	ListNode*  node2 = new ListNode(2);	ListNode*  node3 = new ListNode(3);	ListNode*  node4 = new ListNode(4);	ListNode*  node5 = new ListNode(5);	head->m_pNext  = node1;	node1->m_pNext = node2;	node2->m_pNext = node3;	node3->m_pNext = node4;	node4->m_pNext = node5;	node5->m_pNext = NULL;	int n = GetLength(head);	cout<<"个数为："<<n<<endl;	PrintListBack4(head);	getchar();	return 0;}

C++算法之 倒序输出一个链表